JEE Advanced 2023 — Paper 2 — CHEM-III solution

JEE Advanced2023Paper 2CHEM-III

$H_{2} S$ ( $5$ moles) reacts completely with acidified aqueous potassium permanganate solution. In this reaction, the number of moles of water produced is $x$ , and the number of moles of electrons involved is $y$ . The value of $(x + y)$ is ____.

Your answer

Solution

The balanced equation in acidic medium is

$2 K M n O_{4} + 5 H_{2} S + 3 H_{2} S O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 5 S + 8 H_{2} O$ .

For exactly $5$ moles of $H_{2} S$ (the coefficient in the equation), $8$ moles of $H_{2} O$ are produced, so $x = 8$ .

Electron count. In each half reaction $M n^{+ 7} \to M n^{+ 2}$ gains $5$ electrons. Two moles of $M n O_{4}^{-}$ require $2 \times 5 = 10$ electrons. Equivalently, $5$ moles of sulfide $S^{2 -}$ each lose $2$ electrons giving the same total of $10$ . Therefore $y = 10$ .

$∴ x + y = 8 + 10 = 18$ .

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